Use the result of strlen as length of char* - c++

Question

My intention is to create a char[] from the length of a char* I've already created.
I asked myself why this is not valid:
void copyStringsV2() {
const char* source = "this is my string.";
const int length = strlen(source);
const char* dest[length];
}
The compiler gives me this hint:
Severity Code Description Project File Line Suppression State
Warning C4101 'str_b': unreferenced local variable CStrings
xxx\cstrings.cpp 46
Error C2131 expression did not evaluate to a constant CStrings
xxx\cstrings.cpp 161
Error (active) E0028 expression must have a constant value CStrings
xxx\CStrings.cpp 161
May you can help me out here?

Answer 1

You are trying to declare a variable-length array, which is not a standard feature in C++. A fixed-length array must have its length known at compile-time. That is what the compiler error is complaining about.
If you don't know the length until runtime, you will have to allocate the copied string dynamically instead, such as via new[]:
void copyStringsV2()
{
const char* source = "this is my string.";
const int length = strlen(source);
char* dest = new char[length+1];
strcpy(dest, source);
...
delete[] dest;
}
Or std::vector, so you don't need to use new[]/delete[] directly:
#include <vector>
void copyStringsV2()
{
const char* source = "this is my string.";
const int length = strlen(source);
std::vector<char> dest(length+1);
strcpy(dest.data(), source);
...
}
But, in this case, it would be better to use std::string instead:
#include <string>
void copyStringsV2()
{
const char* source = "this is my string.";
std::string dest = source;
...
}

Answer 2

For starters you should use the type size_t instead of the type int in this declaration
const int length = strlen(source);
^^^
This constant is a constant of the run-time. So you may not use it in the declaration
const char* dest[length];
because here is declared a variable length array and variable length arrays (VLA) is not a standard C++ feature.
Also it is unclear why the type of elements of the array is const char * instead const char.
And moreover a constant object shall be initialized.

Answer 3

My intention is to create a char[]
const char* dest[length];
That's not an array of char. That's an array of pointers to const char.
Also, if you want the array to be able to fit the original null termiated string, you must include the null terminator in the length of the array so that its size is length + 1.
I asked myself why this is not valid:
The compiler gives me this hint:
expression did not evaluate to a constant
May you can help me out here?
The size of an array must be a compile time constant. length is not a constant, therefore it cannot be used as length of an array.
Length of a string cannot be calculated at compile time through a pointer to element of the string. However, if you used a reference for example, and if you used a constexpr function to get the length, and used a constexpr (const works too) variable, then you could use it as the size of an array. There is such function in the C++ standard library:
auto& source = "this is my string.";
constexpr auto length = std::char_traits<char>::length(source);
char dest[length + 1];

Answer 4

why this is not compilable
Because this is C++, and C++ provide a wide variety of tools already1. Your best bet will be std::string. Strings can be copied and passed around with no additional code to write.
#include <string>
int main()
{
const std::string source = "this is my string.";
const std::string dest = source;
// do something with dest
}
1) So you don't need variable length arrays, which are not part of C++.